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Full statement of the question:

Suppose $alpha in mathbbR$, also $f: mathbbR rightarrow mathbbR, f in C^1(mathbbR), and f(0) = 0$

Consider the following ODE:

$partial_t u(t) = f(u(t)), textfor 0 < t < T $

$u(0) = alpha $

where $u:[0,T] rightarrow mathbbR$ is unknown, and the integral equation:

$u(t) = alpha + int^t_0 f(u(s)) textds, 0<t<T $

Prove that there exists $T>0$ such that the above integral equation has a unique solution $u in C([0, T])$. Furthermore, show that $u$ satisfies $u in C^1([0, T])$ and the above ODE.

My Question:

I have shown just about every point of this question, by using the Picard-Lindelof theorem on the following set:

$X = leq K$,

where $K in mathbbR_> 0 , alpha in mathbbR$ and the norm is $|| u || = ^textsup_t in (0,T) |u(t)|$.

However, while this proves there is a $T>0$ such that a solution unique in $X$ exists, I cannot figure how to show such a solution is unique in all of $C([0,T])$. Can someone please inform me how I might show this?

Your work so far has shown that there exists a solution $u$ which is unique in the subset $B_K(alpha)$ (the $K$ ball around the constant function $alpha$ in $C([0,t])$. We want to show this function is in fact unique in the whole space $C([0,t])$.

Suppose $v in C([0,t])$ is also a solution to the IVP, and let $tau = infleft(T cup t: t < T, right)$. Note that, since $v$ is continuous and $v(0) = alpha$, we must have that $tau > 0$.

Observe that,
$$u, v in Y := w in C([0,tau]) : sup_[0,tau] $$
and as a corollary of the Picard-Lindelof contraction argument you performed, you can conclude that $u equiv v$ on $[0,tau]$ (intuitively, we still get uniqueness if we look over a shorter time-scale). If $tau = T$, then $u equiv v$ in $C([0,t])$. Since this is the result we want, our goal now is to show that $tau < T$ gives a contradiction.

Suppose $tau < T$, and let $beta = u(tau)$. By the definition of $tau$, for any $tau < s < T$, $|v(s)| > K geq |u(s)|$. This means that the IVP
$$left{beginarraycc
partial_s w(s) = f(w(s)) &qquad tau < t < T\
w(tau) = u(tau)&
endarrayright.$$
has two solutions in any set $Z = w in C([tau, T^*]) : lVert w - beta rVert < M$ for any $M > 0$ and $T^* > tau$, since $u$ and $v$ instantly separate after time $tau$. But, you proved earlier that we can always find a 'locally unique' solution to this ODE, so this is a contradiction.

In the standard proof of the Picard-Lindelöf existence and uniqueness theorem for IVP in the setup of an autonomous ODE, one first fixes some value for the radius $K$ (so that, at least, the ball of radius $K$ is contained in the domain of $f$). Then one obtains the maximum $M$ of the norm of function values $|f(x)|$ over the ball $xin bar B(α,K)$. Now the condition $MTle K$ provides a first bound for $T$. It ensures by the mean value theorem that any solution $uin C([0,T])$ of the initial value problem, if it exists, stays inside that ball, as
$$
|u(t)-α|=left|int_0^tf(u(s)),dsright|leint_0^t|f(u(s))|,dsle Mtle MTle Ktag1
$$
That is, any function in $C([0,T])$ that solves the IVP is automatically contained in $X$.

Next one further reduces the time interval by adding the restriction $2LTle 1$ where $L$ is the local Lipschitz constant of $f$ on $B(α,K)$ to ensure contractivity of the integral operator in the maximum norm. With everything now fixed you apply the Banach fixed-point theorem that ensures the actual existence of a local solution in $X$ as well as its uniqueness. To repeat, by construction of $T=min(fracKM,frac12L)$, there can be no solutions in $C([0,T])$ that are outside of $X$, see inequality (1).

As a consequence one gets that any solution in $C([0,b))$, $b>T$, when restricted to $[0,T]$ has to coincide with this local solution. Repeated application of that argument can also serve to show that the solution is unique as long as it exists. Because if that were not the case, there would be a first point $t_0$ where two solutions diverge from each other, resp. the last point where they are still the same. However, the IVP with $u(t_0)$ as initial value at $t=t_0$ has a unique local solution, so $t_0$ can not be the minimum under consideration.